使用grep替换单词并在shell中使用sed

Question

I have some 150 files and in them I want to remove this following code:

<SCRIPT LANGUAGE="JavaScript" TYPE="text/javascript" SRC="/height.js"></SCRIPT>

What I'm doing is:

sed -e 's/<SCRIPT LANGUAGE="JavaScript" TYPE="text/javascript" SRC="/height.js"></SCRIPT>/ /' file_names

This doesn't seem to work.

I want to remove this script from all the files in one go. How can I do that?

Answer 1

You have to worry about the slashes in the text you are replacing.

• Either: use ' \\/ ' for each slash,
• Or: cheat and use ' . ' to match any character at the point where the slash should appear.

The alternative exploits the improbability of a file containing the HTML. Theoretically, if you don't like the second alternative, you should also use ' \\. ' at each point where ' . ' appears in the string you're looking at.

sed -e 's/<SCRIPT LANGUAGE="JavaScript" TYPE="text.javascript" SRC=".height.js"><.SCRIPT>/ /' file_names

This is copied from your example and slashes are replaced by dots. However, supplying all the file names on the command line like that will simply write the output as the concatenation of all the edited files to standard output.

Classically, to edit files more or less in situ, you'd write:

tmp=${TMPDIR:-/tmp}/xxx.$$
trap 'rm -f $tmp; exit 1' 0 1 2 3 13 15
for file in ...list...
do
    sed -e '...' $file > $tmp
    mv $tmp $file
done
rm -f $tmp
trap 0

This includes reasonably bullet-proof clean-up of the temporary - it is not perfect. This variant backs up the original before replacing it with the edited version:

tmp=${TMPDIR:-/tmp}/xxx.$$
trap 'rm -f $tmp; exit 1' 0 1 2 3 13 15
for file in ...list...
do
    sed -e '...' $file > $tmp
    mv $file $file.bak
    mv $tmp $file
done
rm -f $tmp
trap 0

With GNU sed , you can use the ' -i ' or ' --in-place ' option to overwrite the files; you can use ' --in-place=.bak ' to create backup copies of each file in file.bak .

Answer 2

You need to escape the special characters with an extra backslash.

Note that the output will also all go to the console. If you want 150 separate output files, you might want to look at the xargs command, something like:

ls -1 | xargs -t -i 'sed -e -i "replace comment" {}'

Be aware that the sed '-i' option will edit the files in place so get your replacement right first and back the files up!

Answer 3

您应该通过管道输出，例如

sed -e 's/<.*SRC="\/height.js".*>//g' < foo.html > foo1.html

Answer 4

You don't have to use the / with the s command:

sed 's|old|new|' files...

If you want to do inline-replace, then the first step is to back up all of your files then issue this command, substitute 'old' and 'new' with strings of your choice:

sed -i .bak 's|old|new|' files...

Because inline replacement is very dangerous, I can't emphasize enough that you must back up all of your files before running the command.

Did I mention that you should back up all of your files ?